Question

Differentiate the given expression ${{\left( \log x \right)}^{x}}+{{x}^{\log x}}$ with respect to $x$.

Answer 1

Hint: Just go on differentiating the both terms separately as it is given with respect to $x$ and the variable in question is $x$, you can directly differentiate by basic properties $d\left( a+b \right)=da+db$ . By this differentiate each term and then add them to get the final answer. Assume each term as a different variable and apply log to find the differentiation of that term significantly.

Complete step-by-step answer:

Given expression in the question for which we need differentiation
${{\left( \log x \right)}^{x}}+{{x}^{\log x}}$
We separate both the terms from the given expression above:
Let us assume that the term ${{\left( \log x \right)}^{x}}$ represented by “a”
Let us assume that the second term ${{x}^{\log x}}$ represented by “b”
Let us assume the total given expression represented by $''y''$
So, now we get that from above assumption we say:
$\begin{align}
  & y={{\left( \log x \right)}^{x}}+{{x}^{\log x}}=a+b \\
 & y=a+b \\
\end{align}$
By differentiating with respect to $x$ on both sides, we get
$\dfrac{dy}{dx}=\dfrac{da}{dx}+\dfrac{db}{dx}..........\left( i \right)$
From our first assumption we say that the value of “a” is:
$a={{\left( \log x \right)}^{x}}$
By applying the logarithm on both sides above, we get:
$\log a=x\log \left( \log x \right)$
Differentiating on both sides with respect to $x$ on both sides, we get:
$\dfrac{1}{a}\dfrac{da}{dx}=\log \left( \log x \right)+\dfrac{x}{\log x}\left( \dfrac{1}{x} \right)$
We use the formulas: $\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x};d\left( uv \right)=udv+vdu$
From above, we get:
$\dfrac{da}{dx}=a\left( \log \left( \log x \right)+\dfrac{1}{\log x} \right)..........\left( ii \right)$
We know
\[b={{x}^{\log x}}\]
By applying logarithm on both sides of equation, we get:
$\log b={{\left( \log x \right)}^{2}}$ . We know $\dfrac{d}{dx}{{x}^{2}}=2x$
By differentiating with respect to $x$ on both sides, we get:
$\dfrac{1}{b}\dfrac{db}{dx}=\dfrac{2\log x}{x}$
By above we can say that value of the term $\dfrac{db}{dx}=2b\dfrac{\log x}{x}.........\left( iii \right)$
By equation (i) of differential, we get the value of $\dfrac{dy}{dx}:$
$\dfrac{dy}{dx}=\dfrac{da}{dx}+\dfrac{db}{dx}$
By substituting equation (ii) and equation (iii) above, we get
$\dfrac{dy}{dx}=a\left( \log \left( \log x \right)+\dfrac{1}{\log x} \right)+2b\dfrac{\log x}{x}$
By substituting a, b values back into equation, we solve it to:
$\dfrac{dy}{dx}={{\left( \log x \right)}^{x}}\left( \log \left( \log x \right)+\dfrac{1}{\log x} \right)+\dfrac{2{{x}^{\log x}}\log x}{x}$
Therefore, this is the final expression.

Note: i) While applying log to a, don’t forget to write $\log \left( \log x \right)$ some may confuse and write $\log x$ . But log should come twice.
(iI) In second term $\log b$ is directly ${{\left( \log x \right)}^{2}}$ you may write $\left( \log x \right)\left( \log x \right)$ for clarity.